 so we have been discussing about this interface effectiveness factor or the effectiveness factors based on our mass transfer steps and the first one what we have taken was the film control diffusion also is very fast diffusion through the force also is very fast so if only film is controlling how do you develop the effectiveness factors and for that we gave the name as interface effectiveness factors okay this is continued we have already done derived one equation yesterday for first order reaction okay so that equation was Eta bar for n equal to 1 first order equal to 1 by 1 plus dA where dA is the Damkohler number okay good and for second order you are going to derive I will give you the final expression as Eta bar equal to of course Eta bar is also R by R actual by R observed okay that is also there Eta bar so then Eta bar equal to 1 by 2 dA Damkohler number then I have 1 plus 4 dA okay to the power of half minus 1 and this whole thing squared so if this number was first step equation number this is 6 then we have this is 7 so you can also derive this one for half order so for half order you have Eta bar equal to 2 plus dA square by 2 then we have 1 minus square root of 1 minus 4 by 2 plus dA square again this is whole square yeah this is this and here everything is under square root this is equation 8 yeah please let me check this because I am copying from a nodes 2 plus dA whole square by 2 correct 1 minus square root of 1 minus 4 by 2 plus dA square whole square all this yeah and the entire thing is under please check this and now I have lots of simple simple problems for my separate test okay one day I can give you this first n equal to 2 next day I can give you n equal to 2 n equal to half next day I can give you n equal to minus 1 okay so many things minus 1 you try what kind of equation you will get good yeah so then when you develop these equations then if you trying to plot that and then see how the effectiveness factors are varying with Damkohler number I have here Eta bar here I have dA right and this dA on log scale it will vary from 0.01 okay this is 0.1110 log scale and this side also it is log scale this is 0.01, 0.1110 good okay so now if I plot this this you know for example n equal to 1 which we understood first so for n equal to 1 you will get first okay like this it may be going like this this is for n equal to 1 anyway I will draw that so for n equal to 2 may be like this for n equal to half may be like this and it is very nice to see for n equal this is just qualitative graphs not exactly may not be to the scale so this is for n equal to minus 1 this equal to half this is for 1 and this is for 2 okay so what is the use of this graph we develop equations and we draw graphs and I think you know you should know why we are drawing graphs it is not that we do not have any work but I think you know there is some meaning you know you can also have lot of conclusions from the graph engineers and scientists always use lots of graphs because one picture can speak 1000 words so that is the reason okay yeah so how do we use this one Swami any idea what is the use of these graphs where did you do your exam take your MS or PhD oh MS where did you do your BTEC DVRIT where did you draw I mean your teachers any time did they draw the graphs on the board and try to explain just asking what what kind of can you remember at least one graph very difficult question that is why memory memory I think because of the computers all points are temporary stored and one semester is the maximum time after one semester all temporary points deleted next semester ready for storing points clean slate yeah clean slate start with so that maximum driving force in there okay how do you see progress of reaction in this graph again I may question where is the conversion here in this graph this is effectiveness factor versus damn color number okay what information you will get from this graph I think all these things are necessary I say that is why I have to take you know every semester 100 classes because I am trying to tell you everything that is my bad luck how do you know from here master's control or not okay so low damn color number somewhere here you have that too all all the orders also you will have effectiveness factor equal to almost one okay so then as damn color number is increasing what is the meaning of damn color number increasing k g is decreasing that is in the denominator know yeah and also please remember that the d i is not universal definition for first order you know k by k by k g a we have written yesterday for first order so that is the first one right in this case but here the definition changes here the definition changes that is why we have the general expression for nth order okay yeah so that is the one and that is the interpretation from the graph but what is the use of the graph apart from interpretation I can now find out what is effectiveness factor for my catalyst what is the information I require for finding out okay to find out how effective my catalyst is what is the information I need to have damn color number how do you get damn color number kinetics that means intrinsic rate I should know and also k g a I should know do you have any equations for k g a lots that depends on what kind of reactor also you are using okay for k g a and k is intrinsic rate constant right but anyway somehow when you know d a which is nothing but k by k g a c b to the power of n minus 1 okay for general nth order so then I will first know what is d a and then I go here for second order point may be point 2 or point 3 okay is the effectiveness factor okay similarly when I go at 1 and then see that may be approximately this is not exact to the scale so that is why again you go here and then you will get almost 1 so then that will give you an idea just by calculate damn color number whether you have any you know the effect the catalyst will be effective or not or are you using the entire surface area for the catalyst or not okay so that is the use of all effectiveness factor versus damn color number and if you go to diffusion through pores there are 2 parameters 1 parameter comes there like d a what is that parameter called you have done already so that is why theory modulus okay so now theory modulus you are supposed to calculate and once you know theory modulus then the graphs will easily allow me to check whether I have effective catalyst or not non effective catalyst right good so that is the reason why we will have the information on this but there is a catch here d a is having intrinsic rate constant and you can never get any information on intrinsic rate constant why in heterogeneous system always whatever you whether you like or not definitely there will be some slight effect of mass transfer so that means that k is not exactly k but it is some apparent k in the presence of mass transfer okay so how do I now get this coordinate without any mass transfer effects and most importantly how do you get that y axis x axis in terms of observables that means observables means measureables in terms of measureables how do you get this coordinate because here k is not observable is not measurable so easily okay what we do normally is that we take very fine powder and then send the gas over the powder and we assume that we do not have any mass transfer limitations and fine powder means it is a vague statement okay what is fine powder for you talcum powder talcum powder is fine powder but I think you know scientifically in terms of size you have to tell no microns or micron or what ya you should have some of these things you know observables in your mind I told you know what is the length of the check piece oh 50 microns you cannot apply you know talcum powder at all skin will come off normally you take like this and then pour no ya so then skin will come off or 50 microns it should be very very fine you know less than 10 1 micron preferably really ya 50 microns if you want to see come to our lab and then we have lot of sand with 50 microns 50 microns 100 microns 200 microns 500 microns we have we will show it to you if you do not believe and you can check that with your talcum powder okay so talcum powder ya so that is why when you say that fine powder it does not really mean anything to us okay and you cannot go to 1 micron 0.5 micron or you know another 1000 of the micron and all that if you go particles also will run off like our dust particles dust particles also around only ya know 1 micron 5 microns 5 microns 10 microns like that okay dust particles if you see your scooter or cycle or car whatever you have you can see you know everyday how much is accumulating also we can try to find out so that is why whatever you do you cannot avoid mass transfer when you are taking the particle that is why carberry has a very beautiful idea let me express this x axis in terms of observable or measurable experimentally experimentally we are supposed to do good measurements correct measurements and then use that coordinate to calculate what will be the eta bar so the next task is to relate one observable to eta bar it is a something equation like this but here you will have observable as the parameter and you can calculate now what is eta bar okay so that is what what we do now you know you know the problem now why we cannot use same problem same problem even with pore diffusion theorem modulus is not observable if you remember for first order theorem modulus because you have done some courses already what is the equation for theorem modulus phi equal to normally many people use phi as theorem modulus time inverse what inverse phi I am asking phi not effectiveness factor you are telling effectiveness factor equal to 1 by 10 h phi as the you know for single pore okay I am not talking about that I am talking about you are trying to recall your memory what you have written the examination right mcp by kn mcp mcp is heat transfer that will not come what is mcp in your language is not specific heat and mass and all that what is mcp because everyone will have our own notation Viola remember okay close l into root k by d l into what is l you said l into oh length okay that is right l into root k by d for first order we call it is for l is here radius of the particle r into square root of k by d again k is intrinsic rate constant why androg you do not sleep normally in the nights you sleep then again I think always you will I mean yawning all the time from the class one thing may be terribly boring to you okay I think that looks because body reactions you cannot stop now okay body reactions I think you cannot stop most of the time okay anyway so the square root of k by d again there also okay is intrinsic rate where you cannot intrinsic rate constant where you cannot measure it so easily so that is why there also we have to do something else to find out what is effectiveness factor so here also we have to do the same thing so then this is this work was done by beautifully carberry so that is why some people call are including me I call that this carberry is observable okay so that carberry is observable if you write that is effectiveness factor effectiveness factor in terms of observables observables okay so we have an equation eta bar equal to r ob that is actual rate observed or actual rate and r b r b is rate based on bulk conditions we are talking about isothermal conditions only here okay but later I will tell you how we can also change to non-hearth thermal conditions good but observable will not change yeah eta bar equal to r ob by r b and we also know that from this equation if I call this okay let me put this one as separate equations so I have r ob equal to eta bar r b r b is the general rate so we can also have this one as eta bar k c b to the power of n or nth order reaction r b or in general this also we can have eta bar k function of c b it need not be always c b to the power of n but it can be you know any form so this is equation 1 okay you know equation 2 good so what we call here is this in this equation r ob r ob equal to this eta bar k we call as k bar c b to the power of n yeah so this is equation 3 where k bar equal to eta bar k okay so now what I do is I just try to manipulate this is equation 4 I will take so this k bar by k g a I just divide and I have here eta bar this will be now because I divided by k g a this side here also I have to divide by k g a yeah and also c b to the power of n minus 1 also I divide by c b to the power of n minus 1 yeah so I have something here you know what is this group damkohler number right so now yeah this is damkohler number so this equation now can be written as eta bar d a correct this we will put as 1 nothing eta bar d a and the other side we have c b to the power of 1 minus n I just taken it up okay so this k bar we will substitute in this equation k bar this is equation number this is 5 so k bar we have from 5 you tell me substitute equation 5 in 3 and tell me what do you get substitute equation 5 in 3 what do you get r o b equal to so much time into k g a into k g a okay so r o b you get as okay once more eta bar eta bar damkohler number as 1 group into c b into k g a so this is the equation so now eta bar d a okay this is equation number 6 eta bar d a as a group I will write now r o b by c b k g a so this is equation 7 and this is an observable that is an observable why that is an observable rachit very seriously looking at that which k where is k there why it is observable I am asking yeah all things are measurable because most of the time for k g a we have lots of correlations whatever type of reactors you take whatever conditions we take and then c b measurable r o b is measurable you are measuring the rate r observed is measurable so that is why this group is called the carabaries observable quantity and this if I now eta bar d a versus eta if I am able to develop equations eta bar d a versus eta eta is the effectiveness factor eta bar d a is nothing but r o b c b k g a so or in other words I have to plot r o b divided by c b k g a versus eta then I measure this go to lab measure this r o b c b k g a and then I can now either calculate or go to the graph and then find out what will be the effectiveness factor now I am sure my effectiveness factor is correct earlier I have always suspicion that how accurate my assumption here for intrinsic rate intrinsic constant right so that is the problem so that is why this has been beautifully solved by carabary to say that yes now you have this observable and use this observable to calculate effectiveness factor for your catalyst this idea I think he got only from the I do not know whether you heard of a criteria called wise prater criteria wise prater W E I S Z wise dash prater P R A T E R again some more grandfathers in chemical engineering that wise prater criteria is used for diffusion through force okay that also we will do that so this will be very similar to that this criteria also this observable that is also an observable in fact there using wise prater criteria you can even find out what should be the diameter of the particle which you have to design that will come under catalyst design apart from scientific design of finding out active sites and trying to impose I mean put some active sites deliberately wherever you want and that is surface science apart from that the macro school design you can that means design of the particle maybe should it be 1 inch or 1 centimeter or 1 millimeter okay why 1 millimeter 1 centimeter 1 1 inch the reason is that we should not have any mass transfer limitation for the particle what will happen if you have mass transfer limitation your surface area is not completely utilized so that is the reason why you have to design this particle okay without any mass transfer limitations most of the time you should be only reaction that is controlling mass transfer should not control the reaction okay no I mean it is like for us also always you know we would like to have as much food as possible whether you work or not that is reaction okay but this is the basic criteria even if you are laid by 1 hour for lunch or lunch or lunch or lunch or I think every student start looking at his watch okay that is the mass transfer is the basic requirement so that is why we would not keep you hungry at all always feed them and try to extract maximum work but that is not possible at least feeding is done all the time okay good so similarly like that only we should design our particle such that the particle should not get angry always there must be sufficient amount of mass that is available the concentrations then of course depending on temperatures and all that it can happily react good yeah so now let us relate this observable in terms of effectiveness factor okay so that is what is the next one okay please write this otherwise I think you may forget Eta bar is an observable full stop okay if Eta bar dA is related to Eta bar analytically if Eta bar dA is related to Eta bar Eta bar dA as one group always Eta bar dA is nothing but observable good if Eta bar dA in the bracket you write is related to Eta bar analytically it would be useful to know about the catalyst to know about the catalyst let us do this in a most general form taking non-isothermal effectiveness taking non-isothermal effectiveness now non-isothermal effects sorry not effectiveness taking non-isothermal effects okay so now we have another title here generalized non-isothermal effectiveness non-isothermal external because we are talking about only film external effectiveness this is the one okay good so let me do that in general we have for any reaction t b is not equal to t s and c b is not equal to c s what is the meaning yeah always there may be some limitations yeah so then Eta bar in this case is defined when you have non-isothermal conditions as actual rate observed okay let me also write that divided by rate if there is no if there are no if there are no mass transfer okay I also write mass and heat transfer effects or limitations that is based on that is rate based on t b slash c b that is based on bulk t b c b good so this Eta bar can be written as k s c s to the power of n divided by k b c b to the power of n yeah of course in normal term this will be r o b r b so this is equation number one okay this is equation number one then I will go here remove all this this derivation again is popularized by carberry what is the spelling of carberry as if you know I am just telling car plus bery car plus bery berries you know what is spelling for car what is for bery bery okay carberry okay is very good very good chemical reaction engineer excellent this derivation you will see how beautifully he has used all basic fundamentals that we describe in chemical engineering in terms of transport wonderful excellent derivation I really enjoy it doing okay yeah so this equation that equation I think maybe I will write here the same thing Eta bar just writing in a slightly different way this is k s by k b c s by c b whole to the power of n you know idea is that this group separated we will find out and this group separated we will find out okay that is the reason why same equation is written in terms of this and k s k b equal to 1 if I have isothermal case okay right so that is why we are going to use that equation equation 2 and yeah this is again just to write same thing I will be writing many times just for I mean for simple algebra so this is nothing but Eta bar k b c b to the power of n of course which is also equal to k s c s to the power of n this is equation number 3 okay so now we have again we have you know either a wire non porous wire or you may have particle non porous particle so for that particle at steady state please remember only film is controlling that first picture you have to see okay only film is controlling at steady state you have r o b rate observed equal to this equation already you know k g a c b minus c s yeah so this is equation equation 4 this already oh no okay at steady state this must be also equal to rate of reaction on the k s c s to the power of n this is equation 4 because at steady state we are equating this is okay r o b this equal to this right sorry what reaction oh okay good so this again r o b I am separating only this k g a by c b minus c s this is equation 5 I will simply divide this r o b by k g a c b yeah equal to c b minus c s by c b which can also be written as 1 minus c s by c b I think you know all this I could have done also in one or two steps but writing more steps so this is equal to 6 okay now probably you would have seen this okay already we find some pattern c s by c b there so now I can write I can write now c s by c b equal to 1 minus r o b by k g a c b which is also equal to this is all one group 1 minus eta bar d a correct right okay so now if I substitute this c s by c b there what do I get substituting equation 7 in 2 substituting equation 7 in 2 what do I get eta bar equal to yeah k s by k b k s b k b into 1 minus eta bar d a this whole group to the power of n yeah so this is the non isothermal effectiveness factor this is yeah for isothermal case k s equal to k b then eta bar equal to 1 minus eta bar d a equal to the power of n n and eta both should not look same n has I mean eta has more tail this is also eta bar d a okay so that means now you can see very easily now I can calculate this one for n equal to 1 n equal to 2 n equal to half okay same thing what we have done but that equation is different this equation is different and here I am very sure what is eta bar d a which I measure right so measure in the sense you know I have all this parameters we can also plot this graph now as eta bar versus okay yeah eta bar versus eta bar d a or eta bar versus r o b divided by c b k g a I am now more confident for isothermal case we have a relationship now eta bar in terms of observables observables are nothing but your rate observed c b measurable k g a measurable or you have the equations good so now how do I extend this for this will plot later for n equal to 1 n equal to 2 n equal to half and all that okay this equation plotting eta bar versus observable good so that we will do later but now here k s what is the equation in terms of temperature arhenius equation okay again we have the problem in arhenius equation that k is supposed to be intrinsic reaction rate constant okay so now we have to somehow eliminate that you know the temperature that k means when I say that arhenius equation k equal to k not or k s equal to k not e power minus d a what is that t where is the operating condition on the surface it is not t it is t s okay k s is k not into e power minus e by r t s but again the t s is not measurable for right like c s is not measurable t s also is not measurable so that is the reason why I cannot use this equation straight so I have to do something else that again I have to make everything in terms of observable everything in terms of measureables that I think beautifully done by carberry okay so what we do now here is this k s by k b we will try to substitute for non-isothermal k s and we have an equation for you know k s by if I write k s and k b in terms of arhenius equation so that equations can be this is equation 9 k s by k b time up for you okay this can be written as a into e power minus e by r t s divided by a a will not change e power minus e by e also will not change r also will not change only t b t s right so this equation can be written as e power minus e by r t b okay e pi r t b and one more step I will write here this is t b t s minus 1 yeah okay that is fine good so of course same thing we can also write k b in terms of some dimensionless numbers that is exponential epsilon not 1 by small t minus 1 okay of course if I want I can also put another bracket there where epsilon not equal to you have come across that number where epsilon not equal to absence what is that what do you say e by r t b yeah you know the name of that assume already b bold arhenius constant arhenius number that is arhenius number and t equal to small t t s by t b so numbers if I put this is 10 this is 11 and of course those 2 I think that is okay no numbers there so now this k s by equation 11 I can substitute in equation 8 okay substituting equation 11 in equation 8 now you will have eta bar equal to exponential minus epsilon or epsilon not okay I think in for epsilon not I can put epsilon b why based on t b n okay based on t b okay so epsilon b 1 by t minus 1 this equation and okay this multiplied by 1 minus eta bar d a this also whole to the power of n so this is equation number 12 okay is it useful now this is observable this is arhenius number observable do I know a priori what is arhenius number yes you know definitely you know yeah so observable and t not observable so that is why now you have to go to heat and mass balance around the pellet and then try to relate that in terms of measurable really wonderfully done I think that I cannot do now because I think you have you have to run for that class even though I have time you do not have time okay so what we do here is that the particle you take under steady state conditions like we have written at steady state conditions now this is the mass balance now we use heat balance what is the heat balance there what is the heat balance yeah where is convection where is convection are you taking a porous particle by the way first of all so then why porous and all that way the amount of heat generated by the equation on the surface must be equal to heat supplied to the surface if it is endothermic or heat brought out from the surface if it is exothermic so this particle will be hot if it is exothermic but outside temperature is less so naturally heat flows in this direction and if it is endothermic reaction H B sorry T B will be higher endothermic you know on the surface you will have less temperature so heat flows in the other way so always exothermic is exiting we will take exothermic reaction okay good I think that we will do that balance and now you will get in terms of heat transfer coefficient and heat of reaction and all that so you will have another parameter called beta in terms of observable and everything can be expressed in terms of T B like C B we have expressed so that is how you finally get Eta bar this equation everything in terms of observable that means for T S by T B you have to write an equation in terms of observable it is a wonderful design everything you know you heard of what is J H J D factors and all that those things also will come here yeah do you know what is the relationship between J D and J H yeah simplest thing is I think you know that is under certain conditions what she said is right J D equal to J H but under some other conditions J D equal to point 7 J H or so okay but anyway we take J D is J H and I may give in the examination J D equal to point 7 J H okay good particularly for if you take only a packet bed reactor you may have J D equal to J H point 7 J H okay there is some you see all these criteria you have already learnt in transport phenomena momentum transfer and all that okay so that is why I think all these things we are beautifully going to use to solve this problem okay now you run thank you